节: 应用:流体力量 | 6.9 应用:流体力量

章节大纲

In the branch of physics known as fluid mechanics, some problems require the determination of forces or pressures acting on an object submerged in a static fluid (i.e., a fluid at rest). Problems can involve the force or pressure exerted by a column of fluid over an object, or the force or pressure exerted on the sides of a dam or storage tank that encloses the fluid. Do you know which of the two types of problems is easier to solve and why?
::在被称为流体力学的物理分支中,有些问题需要确定在静态流体(即静态流体)淹没的物体上产生的力或压力(即静态流体),问题可能涉及一列流体对物体施加的力或压力,或对淹没流体的大坝或储油罐两侧施加的力或压力。你知道这两种问题中哪一种比较容易解决,为什么?

Fluid Forces
::流流力量

You have probably studied that pressure is defined as force acting over an area $\begin{align*}P=\frac{F}{A}\end{align*}$ which has the units of Pascals $\begin{align*}(Pa)\end{align*}$ , or Newtons per meter squared, $\begin{align*}Pa=\frac{N}{m^2}.\end{align*}$
::你可能已经研究过压力的定义是力在P=FA区域上行动其单位为Pascals(Pa), 或牛顿每平方公尺,Pa=Nm2。

In the study of fluids, such as water pressure on a dam or water pressure in the ocean at a depth $\begin{align*}h\end{align*}$ an equivalent formula can be used called the liquid pressure $\begin{align*}P\end{align*}$ at depth $\begin{align*}h\end{align*}$ :
::在研究液体时,例如水坝的水压或深海的水压,在深度 h 可使用一个等量的公式,称为深度 h 的液体压力P:

$\begin{align*}P=wh=\rho gh\end{align*}$ , where $\begin{align*}w\end{align*}$ is weight density (the weight of the fluid column per unit volume).
::P=whgh,重量密度(每单位体积液列重量)。

$\begin{align*}\rho\end{align*}$ is mass density of the fluid (mass of fluid per unit volume), and $\begin{align*}g\end{align*}$ is acceleration due to gravity ( $\begin{align*}g=9.8 \ m/s^2\end{align*}$ on Earth).
::是液体的质量密度(每单位体积的液体质量),而g是重力加速度(g=9.8 m/s2在地球上)。

For example, if you are underwater in a pool, the pressure of the water on your body can be calculated by determining the weight density of the column of water above you times your depth.
::例如,如果在水下水池中,通过确定上方水柱的重量密度乘以深度,就可以计算出身体上水的压力。

What is the fluid pressure (excluding the air pressure) and fluid force on the top of a flat circular plate of radius 3 meters that is submerged horizontally at a depth of 10 meters ?
::半径3米的平面圆形板顶部的液压(不包括气压)和液压是什么?

The density of water is $\begin{align*}\rho=1000 \ kg/m^3\end{align*}$ . Then
::水的密度为1 000千克/立方米。

$\begin{aligned} P & = ρ g h \\ = (1000) (9.8) (10) \\ = 98000 P a . \end{aligned}$
$\begin{align*}P&=\rho gh\\ &=(1000)(9.8)(10)\\ &=98000\ Pa.\end{align*}$
::Pgh=(1000)(9.8)(10)=98000帕。

Since the force is $\begin{align*}F=PA\end{align*}$ , then
::既然部队是F=PA,那么

$\begin{aligned} F & = P A \\ = P \cdot π r^{2} \\ = (98000) (π) (3)^{2} \\ = 2.77 \times 10^{6} N . \end{aligned}$
$\begin{align*}F &=PA\\ &=P\cdot \pi r^2\\ &=(98000)(\pi)(3)^2\\ &=2.77\times 10^6\ N.\end{align*}$
::F=PA=Pr2=(98000)(__)(3)2=2.77×106 N。

As you can see, it is easy to calculate the fluid force on a horizontal surface because each point on the surface is at the same depth. The problem becomes more complicated when we want to calculate the fluid force or pressure if the surface is vertical rather than horizontal. In this situation, the pressure is not constant at every point on the surface because the depth is not constant at each point. To find the fluid force or pressure on a vertical surface we must use calculus.

::正如您所看到的, 计算水平表面的流体力很容易, 因为表面的每个点都是在同一深度上。当我们想要计算表层为垂直而不是水平的流体力或压力时, 问题就变得更加复杂。在这种情况下, 压力在表层的每个点上并不是恒定的, 因为每个点的深度不是恒定的。要找到垂直表面的流体力或压力, 我们就必须使用计分法。

The Fluid Force on a Vertical Surface
::垂直表面的流流力

Suppose a flat surface is submerged vertically in a fluid of weight density $\begin{align*}w\end{align*}$ and the submerged portion of the surface extends from $\begin{align*}x=a\end{align*}$ to $\begin{align*}x=b\end{align*}$ along the vertical $\begin{align*}x\end{align*}$ -axis, whose positive direction is taken as downward. If $\begin{align*}L(x)\end{align*}$ is the width of the surface and $\begin{align*}h(x)\end{align*}$ is the depth of point $\begin{align*}x\end{align*}$ , then the fluid force $\begin{align*}F\end{align*}$ is defined as
::假设平平面被垂直水淹在重密度的液体中,而表下沉部分从x=a到x=b沿垂直x轴向下倾斜,其正向为向下向。如果L(x)是表面宽度,h(x)是点x的深度,那么液力F的定义是:

$\begin{align*}F=\int\limits_{a}^{b}wh(x)L(x)dx.\end{align*}$
::Fabwh(x)L(x)dx。

This formulation of the force on the side of a vertical surface works because liquids are virtually incompressible, and the pressure on any part of an object (horizontal or vertical) at the same depth is the same. The French scientist Blaise Pascal (1623-1662) recognized that the pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel. This result is called Pascal’s principle .
::法国科学家布莱斯·帕斯卡尔(Blaise Pascal (1623-1662))承认,对封闭液体施加的压力不波及装有容器的液体和墙壁的每一部分,因此,这种压力在垂直表面工程的侧面形成,因为液体几乎无法压缩,同一深度的物体(横向或垂直)的任何部分的压力是一样的。法国科学家布莱斯·帕斯卡尔(1623-1662)承认,对封闭液体施加的压力不波及容器的液体和墙壁。这一结果被称为帕斯卡尔的原则。

A perfect example of a vertical surface is the face of a dam. We can picture it as a rectangle of a certain height and certain width. Let the height of the dam be 100 meters and of width of 300 meters. Find the total fluid force exerted on the face if the top of the dam is level with the water surface.
::垂直表面的完美例子就是水坝的表面。我们可以把它想象成一定高度和一定宽度的矩形。让大坝的高度为100米,宽度为300米。如果大坝的顶部与水面保持平坦, 就可以找到面部的液压总量。

Let $\begin{align*}x\end{align*}$ equal the depth of the water. At the depth $\begin{align*}x\end{align*}$ on the dam, the width of the dam is $\begin{align*}L(x)=300\ m\end{align*}$ and the depth is $\begin{align*}h(x)=x\end{align*}$ meters. The weight density of water is
::让xx等于水的深度。在大坝的深度x时,大坝的宽度为L(x)=300米,深度为h(x)=x米。水的重量密度为:

$\begin{aligned} w_{w a t e r} & = ρ g \\ = (1000) (9.8) \\ = 9800 N / m^{2} . \end{aligned}$
$\begin{align*}w_{water} &=\rho g\\ &=(1000)(9.8)\\ &=9800\ N/m^2.\end{align*}$
::wwaterg=(1000)(9.8)=9800 N/m2。

Using the fluid force formula above,
::使用上面的流力配方

$\begin{aligned} F & = \int_{a}^{b} w h (x) L (x) d x \\ = \int_{0}^{100} (9800) (x) (300) d x \\ = 2.94 \times 10^{6} \int_{0}^{100} x d x \\ = 2.94 \times 10^{6} {[\frac{x^{2}}{2}]}_{0}^{100} \\ = 1.47 \times 10^{10} N . \end{aligned}$
$\begin{align*}F &=\int\limits_{a}^{b}wh(x)L(x)dx\\ &=\int\limits_{0}^{100}(9800)(x)(300)dx\\ &=2.94\times 10^6\int\limits_{0}^{100}xdx\\ &=2.94\times 10^6\left[\frac{x^2}{2}\right]^{100}_0\\ &=1.47\times 10^{10}N.\end{align*}$
::Fabwh(x)L(x)dx=0100(9800(x)(300)dx=2.94×106}0100xx=2.94×106}[x22]0100=1.47×1010N。

The total force on the vertical dam wall is $\begin{align*}1.47\times 10^{10}\ N\end{align*}$ .
::垂直堤坝墙上的总力为1.47×1010 N。

Examples
::实例

Example 1
::例1

Earlier, you were asked which is easier to solve: (1) finding the force or pressure exerted by a column of fluid over an object, or (2) finding the force or pressure exerted on the sides of an enclosure that that contains the fluid, like a dam or storage tank. If you answered (1) you would be correct. Finding the pressure or force on the side of an enclosure involves accounting for the variation of force and pressure with depth through a definite integral. Finding the force or pressure exerted by a column of fluid over an object only requires knowledge of the depth and weight density of the fluid.
::早些时候,有人问您哪一种比较容易解答1) 发现一列液体对一个物体施加的力或压力,或(2) 发现装有液体的封闭物的两侧所施加的力或压力,如大坝或储油罐。如果你回答(1) 正确的话,你就会正确。在封闭物一侧所施加的压力或力量涉及通过一个明确的整体体来计算力量和压力的变异。在物体上找到一列液体所施加的力或压力,只需要了解液体的深度和重量密度。

Example 2
::例2

A storage tank in the shape of a right elliptical cylinder of width 10 meters and height 6 meters holds some fluid that has a mass density of $\begin{align*}\rho=680\ kg/m^3\end{align*}$ . The fluid fills the tank to a height of 3 meters as shown in the figure. What is the force exerted by the liquid on each end of the tank?
::如图所示,储油罐的形状是右椭圆圆柱,宽10米,高度6米,装有某种液体,质量密度为680千克/立方米。液体填充罐体高度为3米,如图所示。液体对罐体每一端的强度是多少?

Since the mass density is $\begin{align*}\rho=680\ kg/m^3\end{align*}$ , the weight density is $\begin{align*}w=\rho g=6664\ N/m^3\end{align*}$ .
::由于质量密度为680千克/立方米,重量密度为 wg=6664N/m3。

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$\begin{aligned} F & = \int_{c}^{d} w h (y) L (y) d y \\ = w \int_{0}^{- 3} y (2 \sqrt{5^{2} - \frac{5^{2}}{3^{2}} y^{2}}) d y & \dots Depth is measured along the negative y -axis. \\ = w \int_{0}^{- 3} y (2 \cdot \frac{5}{3} \sqrt{3^{2} - y^{2}}) d y \\ = w \frac{10}{3} \int_{0}^{- 3} y (\sqrt{3^{2} - y^{2}}) d y \end{aligned}$
$\begin{align*}F &=\int\limits_{c}^{d}wh(y)L(y)dy\\ &=w\int\limits_{0}^{-3}y\left(2\sqrt{5^2-\frac{5^2}{3^2}y^2}\right)dy &&\ldots \text{Depth is measured along the negative }y\text{-axis.}\\ &=w\int\limits_{0}^{-3}y\left(2\cdot\frac{5}{3}\sqrt{3^{2}-y^{2}}\right)dy\\ &=w\frac{10}{3}\int\limits_{0}^{-3}y\left(\sqrt{3^2-y^2}\right)dy \end{align*}$
::FcdwhLdy=w0-3y(252-5232y2y2y)dy.Deph沿着负 Y-轴测量。=w0-3y(255332-y2)dy=w1030-3y(32-y2)dy。

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$\begin{aligned} = w \frac{10}{3} (- \frac{1}{2}) \int_{9}^{0} (\sqrt{u}) d u \dots Use substitution: u = 9 - y^{2} and d u = - 2 y d y . \\ = - w \frac{5}{3} [- \int_{0}^{9} (\sqrt{u}) d u] \dots Use the definition: \int_{a}^{b} f (y) d y = - \int_{b}^{a} f (y) d y when a < b . \\ = w \frac{5}{3} {[\frac{2}{3} u^{\frac{3}{2}}]}_{0}^{9} \\ = w \frac{10}{9} [9^{\frac{3}{2}}] \\ F & = 30 w \dots F = 199, 920 N . \end{aligned}$
$\begin{align*}&=w\frac{10}{3}\left(-\frac{1}{2}\right)\int\limits_{9}^{0}\left(\sqrt{u}\right)du \quad \ldots\text{Use substitution:} \ u=9-y^2 \text{ and } du=-2ydy.\\ &=-w\frac{5}{3}\left[-\int\limits_{0}^{9}(\sqrt{u})du\right] \ \quad\quad \ldots \text{Use the definition:} \ \int\limits_{a}^{b}f(y)dy=-\int\limits_{b}^{a}f(y) dy \text{ when } a < b .\\ &=w\frac{5}{3}\left[\frac{2}{3}u^{\frac{3}{2}}\right]_0^9\\ &=w\frac{10}{9}\left[9^{\frac{3}{2}}\right]\\ F &=30 \ w\ \ \quad\qquad\qquad \qquad\qquad\ldots\ F=199,920 \ N.\end{align*}$
::=w103(-12) 90(u)du... 使用替代: u=9-y2 和 du2ydy. w53 [09(u)du] 使用定义: @abfdybaf y) 当 a <b.=w53[23u32]09=w109[932]F=30 w...F=199,920 N.

The force on either side of the tank is 199,920 Newtons.
::坦克两侧的力量是199 920牛顿

Example 3
::例3

What is the total pressure experienced by a diver in a swimming pool at a depth of 2 meters ?
::潜水员在2米深深的游泳池里所经历的总压力是多少?

First we calculate the fluid pressure the water exerts on the diver at a depth of 2 meters:
::首先我们计算水对潜水员的液体压力深度为2米

$\begin{align*}P=\rho gh\end{align*}$
::

The density of water is $\begin{align*}\rho=1000\ kg/m^3\end{align*}$ , thus
::因此,水的密度为1 000千克/立方米。

$\begin{aligned} P & = (1000) (9.8) (2) \\ = 19600 P a . \end{aligned}$
$\begin{align*}P &=(1000)(9.8)(2)\\ &=19600\ Pa.\end{align*}$
::P=(1000)(9.8)(2)=1600帕。

The total pressure on the diver is the pressure due to the water plus the atmospheric pressure. If we assume that the diver is located at sea-level, then the atmospheric pressure at sea level is about $\begin{align*}10^5\ Pa\end{align*}$ . Thus the total pressure on the diver is
::潜水员的总压力是水加上大气压力造成的压力。如果我们假设潜水员位于海平面,那么海平面的大气压力大约是105帕。因此,潜水员的总压力是105帕。

$\begin{aligned} P_{t o t a l} & = P_{w a t e r} + P_{a t m} \\ = 19600 + 10^{5} \\ = 119600 \\ = 1.196 \times 10^{5} P a . \end{aligned}$
$\begin{align*}P_{total} &=P_{water}+P_{atm}\\ &=19600+10^5\\ &=119600\\ &=1.196 \times 10^5 \ Pa.\end{align*}$
::P总计=Pwater+Patm=1600+105=119600=1.196×105帕。

Review
::回顾

A rectangular swimming pool is 8 meters deep, 10 meters wide, and 16 meters long. Calculate the fluid force on the bottom of the pool when it is filled completely with water.
::矩形游泳池深8米,宽10米,长16米。计算游泳池底部的水力,当水完全填满时。

For the above problem, calculate the fluid force on each 8 m by 10 m end of the pool when it is filled completely with water.
::对于上述问题,当池水完全装满水时,计算水池端每8米乘10米时的流体力。

For the above 8 m by 10 m by 16 m pool configuration, calculate the fluid force on a 0.5 m by 1 m rectangular plate held parallel to the bottom of the pool at a depth of 5 meters.
::对于8米以上乘10米乘16米的水池配置,计算一个0.5米乘1米的长方形板的液体力,该板与水池底平行,深5米。

For the above 8 m by 10 m by 16 m pool configuration, calculate the fluid force on a 0.5 m by 1 m rectangular plate held at and angle of 45 degrees with respect to the bottom of the pool, and with the top end of the plate (of 0.5 m width) held at a depth of 5 meters.
::对于8米以上乘10米乘16米的池子配置,计算在0.5米乘1米的矩形板块上,与池子底部保持45度和角度为45度的液体力,并计算在5米深处保持板顶部(宽度为0.5米)。

For the above 8 m by 10 m by 16 m pool configuration, calculate the fluid force on a 0.5 m by 1 m rectangular plate held at and angle of 45 degrees with respect to the bottom of the pool, if the 1 m end of the plate is held at a depth of 5 meters.
::对于8米以上乘10米乘16米的池子配置,计算0.5米乘1米的矩形板的液体力,如果板子的1米端保持在5米深处,则对池底保持45度的角和0.5米长的长方形板的液体力。

The bottom of a rectangular swimming pool, whose bottom is an inclined plane, is shown below. (Note: the plane of the top of the pool is parallel to the ground; the bottom of the pool is not.) Calculate the fluid force on the bottom of the pool when it is filled completely with water.
::矩形游泳池底部,其底部为倾斜平面,如下所示。 (注:池顶的平面与地面平行;池底不平行。 )计算池底部的液态,当池底完全填满水时。

The vertical face of a dam has a height of 50 meters and of width of 600 meters. Find the total fluid force exerted on the face if the top of the dam is level with the water surface.
::大坝的垂直面部高度为50米,宽度为600米。如果大坝的顶部与水面保持平整,可以发现面部的液压总量。

A storage tank in the shape of a circular cylinder of diameter 10 meters holds some fluid that has a mass density of $\begin{align*}\rho=680 \ kg/m^3 \end{align*}$ . The fluid fills the tank to a height of 3 meters. What is the force exerted by the liquid on each end of the tank?
::以直径为10米的圆圆圆圆柱形的储油罐含有一些液体,其质量密度为680千克/立方米。液体填充罐体的高度为3米。液体对罐体每一端的强度是多少?

A storage tank in the shape of a right elliptical cylinder of width 6 meters and height 4 meters holds some fluid that has a mass density of $\begin{align*}\rho=1000 \ kg/m^3 \end{align*}$ . The fluid fills the tank to a height of 1 meter. What is the force exerted by the liquid on each end of the tank?
::以右椭圆圆筒形状的储油罐,宽6米,高度4米,装有质量密度为1 000公斤/立方米的液体。液体填满罐体高度为1米。液体对罐体每一端的强度是多少?

A flat surface is submerged vertically in a fluid of weight density $\begin{align*}w\end{align*}$ . If the weight density $\begin{align*}w\end{align*}$ is doubled, is the force on the plate also doubled? Explain.
::平面被垂直淹没在重量密度的流体中。如果重量密度加倍, 板块上的力量是否也加倍? 解释。

Suppose that a rectangle that’s a meter long and half a meter wide is standing straight on the bottom of a five meter pool. What is the force exerted by the fluid on one side of the rectangle? The mass-density of water is $\begin{align*}\rho=1000 \ kg/m^3 \end{align*}$ .
::假设一个长一米半米宽的矩形正直站在一个5米宽的游泳池的底部。矩形一边的流体施加了什么样的力? 水的质量密度是1 000千克/立方米。

Suppose that a dish whose radius is two meters is standing upright on the bottom of a ten meter pool. What is the integral that expresses the force exerted by the fluid on one side of the plate? The mass-density of water is $\begin{align*}1000\ kg/m^3\end{align*}$ .
::假设一个半径为两米的盘子站在10米水池底部。什么是显示板上一面液体所施加的力力的元件?水的质量密度是1000公斤/立方米。

Suppose that an equilateral triangular plate that is a meter high is standing straight on the bottom of a fifteen meter pool, resting on its point. What is the force exerted by the fluid on one side of the triangle? The mass-density of water is $\begin{align*}1000\ kg/m^3\end{align*}$ .
::假设一米高的等边三角板正直站在15米水池底部, 停留在它的位置上。三角形一边的液体所施加的强度是多少? 水的质量密度是1000千克/立方米。

Suppose that a square whose side is a meter long is standing straight on the bottom of a eighteen meter pool. What is the force exerted by the fluid on one side of the square? The weight-density of water is 62.
::假设一面长一米长的方形正直站在十八米水池底部。方形一面的液体所施加的强度是多少? 水的重量密度是62。

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

章节大纲

Fluid Forces ::流流力量

The Fluid Force on a Vertical Surface ::垂直表面的流流力

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Review ::回顾

Review (Answers) ::回顾(答复)

Fluid Forces
::流流力量

The Fluid Force on a Vertical Surface
::垂直表面的流流力

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Review
::回顾

Review (Answers)
::回顾(答复)